How to solve x^n = k?

x^n = k

I'm supposed to find all complex solutions. I know that for x^n = 1, it's just the nth roots of unity, but how would you solve it if the RHS wasn't 1

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  • Anonymous
    10 years ago
    Favorite Answer

    It depends on what k can be. Is k always a positive integer? Is k always a negative integer? Can k be a complex number? We will need to consider all of these cases separately.

    Case (i): k is a positive integer.

    Since 1 = cos 0 + i sin 0, we see that, in polar form:

    k = k(cos 0 + i sin 0).

    By DeMovire's Theorem:

    x = k^(1/n)

    = [k(cos 0 + i sin 0)]^(1/n)

    = k^(1/n) * [cos(2πk/n) + i sin(2πk/n)], where k = 0, 1, 2, ..., k-1.

    Case (b): k is a negative integer.

    Since -1 = cos π + i sin π, we see that:

    k = |k|(cos π + i sin π).

    Then, by DeMovire's Theorem:

    x = k^(1/n)

    = [|k|(cos π + i sin π)]^(1/n)

    = |k|^(1/n) * {cos[(π + 2πk)/n] + i sin[(π + 2πk)/n]}, where k = 0, 1, 2, ..., k-1.

    The last case can be done by just using DeMovire's Theorem on the complex number.

    I hope this helps!

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